Suma diviziorilor unui numar | Metode eficiente

//Metoda 1:

int main()
{
    int x,div;
    long long S=0;

    cin>>x;

    for (div=1; div*div < x; div++)
        if (x % div == 0)
            S = S + div + x/div;

    if ( div * div == x)
        S= S + div;

    cout<<S;

    return 0;
}



//Metoda 2: (de prosti, facuta de mine: faci produsul sumelor geometrice a unor progresii date de factorii primi ai
// numarului ). Metoda este, totusi, eficienta.

#include <iostream>
#include <cmath>

using namespace std;

long long suma_geometrica(long long baza, long long exp)
{
    if (baza==1)
        return (exp+1);

    long long s= (1- pow(baza,exp)*baza)/(1-baza);
    return s;
}

long long suma_divizori(long long x)
{
    long long div=2,p,S=1;

    while (x != 1)
    {
        p=0;

        while (x % div == 0)
        {
            x=x/div;
            p++;
        }



        if (p) {
                S=suma_geometrica(div,p)*S;
                //cout<<div<<" la puterea "<<p<<" ";
            }

        div++;
    }

    return S;

}

int main()
{
  long long x;
  cin>>x;

  cout<<suma_divizori(x);

  return 0;

}